Question

In standard Young's double-slit experiment, visible light $\lambda \in \left(350 nm \text{, } 750 nm\right)$ is used in both the slits. Distance between the slits is $2mm$ , and the distance of the screen from the slits is $1m$ . A thin glass slab $\left(\mu = \text{1.5}\right)$ of thickness $6\mu \text{m}$ is placed in front of slit $S_{2}$ . Which among the following wavelengths is missing at point $P,$ which is directly in front of slit $S_{2}$ ?(Neglect dispersion of light)

• A. $400 \, nm$
• B. $500 \, nm$
• C. $600nm$
• D. $700nm$

Answer 1

Answer: (A)

Net path difference $= \left(\mu - 1\right) \text{t} - \frac{\text{yd}}{\text{D}}$
$=\left(\frac{3}{2} - 1\right)6 \, \mu \text{m}-\frac{\left(\text{d} / 2\right) \text{d}}{\text{D}}=1 \, \mu \text{m}$
For Destructive interference
$\Delta \text{x}=1 \, \mu \text{m}=\left(2 \text{n} + 1\right)\frac{\lambda }{2}\Rightarrow \lambda =\frac{2}{\left(2 \text{n} + 1\right)} \, \mu \text{m}$
$\lambda = \frac{\text{2000 nm}}{\left(2 + 1\right)} = \frac{\text{2000}}{3}$ $\left(\right. \, n=1 \, \left.\right)$
$\lambda =\frac{\text{2000 nm}}{\left(4 + 1\right)}=\frac{\text{2000}}{5}\text{nm}=\text{400 nm}$ $\left(\right. \, n=2 \, \left.\right)$